The Stacks Project

Tag 0CLQ

86.34. Valuative criterion for second diagonal

The converse statement has already been proved in Lemma 86.33.2. The criterion itself is the following.

Lemma 86.34.1. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. If $\Delta_f$ is quasi-separated and if for every diagram (86.33.1.1) and choice of $\gamma$ as in Definition 86.33.1 the category of dotted arrows is a setoid, then $\Delta_f$ is separated.

Proof. We are going to write out a detailed proof, but we strongly urge the reader to find their own proof, inspired by reading the argument given in the proof of Lemma 86.33.2.

Assume $\Delta_f$ is quasi-separated and for every diagram (86.33.1.1) and choice of $\gamma$ as in Definition 86.33.1 the category of dotted arrows is a setoid. By Lemma 86.6.1 it suffices to show that $e : \mathcal{X} \to \mathcal{I}_{\mathcal{X}/\mathcal{Y}}$ is a closed immersion. By Lemma 86.6.4 it in fact suffices to show that $e = \Delta_{f, 2}$ is universally closed. Either of these lemmas tells us that $e = \Delta_{f, 2}$ is quasi-compact by our assumption that $\Delta_f$ is quasi-separated.

In this paragraph we will show that $e$ satisfies the existence part of the valuative criterion. Consider a $2$-commutative solid diagram $$\xymatrix{ \mathop{\rm Spec}(K) \ar[r]_x \ar[d]_j & \mathcal{X} \ar[d]^e \\ \mathop{\rm Spec}(A) \ar[r]^{(a, \theta)} & \mathcal{I}_{\mathcal{X}/\mathcal{Y}} }$$ and let $\alpha : (a, \theta) \circ j \to e \circ x$ be any $2$-morphism witnessing the $2$-commutativity of the diagram (we use $\alpha$ instead of the letter $\gamma$ used in Definition 86.33.1). Note that $f \circ \theta = \text{id}$; we will use this below. Observe that $e \circ x = (x, \text{id}_x)$ and $(a, \theta) \circ j = (a \circ j, \theta \star \text{id}_j)$. Thus we see that $\alpha$ is a $2$-arrow $\alpha : a \circ j \to x$ compatible with $\theta \star \text{id}_j$ and $\text{id}_x$. Set $y = f \circ x$ and $\beta = \text{id}_{f \circ a}$. Reading the arguments given in the proof of Lemma 86.33.2 backwards, we see that $\theta$ is an automorphism of the dotted arrow $(a, \alpha, \beta)$ with $$\gamma : y \circ j \to f \circ x \quad\text{equal to}\quad \text{id}_f \star \alpha : f \circ a \circ j \to f \circ x$$ On the other hand, $\text{id}_a$ is an automorphism too, hence we conclude $\theta = \text{id}_a$ from the assumption on $f$. Then we can take as dotted arrow for the displayed diagram above the morphism $a : \mathop{\rm Spec}(A) \to \mathcal{X}$ with $2$-morphisms $(a, \text{id}_a) \circ j \to (x, \text{id}_x)$ given by $\alpha$ and $(a, \theta) \to e \circ a$ given by $\text{id}_a$.

By Lemma 86.33.11 any base change of $e$ satisfies the existence part of the valuative criterion. Since $e$ is representable by algebraic spaces, it suffices to show that $e$ is universally closed after a base change by a morphism $I \to \mathcal{I}_{\mathcal{X}/\mathcal{Y}}$ which is surjective and smooth and with $I$ an algebraic space (see Properties of Stacks, Section 85.3). This base change $e' : X' \to I'$ is a quasi-compact morphism of algebraic spaces which satisfies the existence part of the valuative criterion and hence is universally closed by Morphisms of Spaces, Lemma 55.41.1. $\square$

The code snippet corresponding to this tag is a part of the file stacks-morphisms.tex and is located in lines 7736–7828 (see updates for more information).

\section{Valuative criterion for second diagonal}
\label{section-valuative-second}

\noindent
The converse statement has already been proved in
Lemma \ref{lemma-cat-dotted-arrows}.
The criterion itself is the following.

\begin{lemma}
\label{lemma-setoids-and-diagonal}
Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks.
If $\Delta_f$ is quasi-separated and if for every diagram
(\ref{equation-diagram}) and choice of $\gamma$ as in
Definition \ref{definition-fill-in-diagram}
the category of dotted arrows
is a setoid, then $\Delta_f$ is separated.
\end{lemma}

\begin{proof}
We are going to write out a detailed proof, but we strongly urge the
reader to find their own proof, inspired by reading the argument
given in the proof of Lemma \ref{lemma-cat-dotted-arrows}.

\medskip\noindent
Assume $\Delta_f$ is quasi-separated and for every diagram
(\ref{equation-diagram}) and choice of $\gamma$ as in
Definition \ref{definition-fill-in-diagram}
the category of dotted arrows is a setoid.
By Lemma \ref{lemma-diagonal-diagonal} it suffices to show that
$e : \mathcal{X} \to \mathcal{I}_{\mathcal{X}/\mathcal{Y}}$
is a closed immersion. By
Lemma \ref{lemma-first-diagonal-separated-second-diagonal-closed}
it in fact suffices to show that $e = \Delta_{f, 2}$ is
universally closed.
Either of these lemmas tells us that $e = \Delta_{f, 2}$ is quasi-compact
by our assumption that $\Delta_f$ is quasi-separated.

\medskip\noindent
In this paragraph we will show that $e$ satisfies the existence
part of the valuative criterion. Consider a $2$-commutative solid diagram
$$\xymatrix{ \Spec(K) \ar[r]_x \ar[d]_j & \mathcal{X} \ar[d]^e \\ \Spec(A) \ar[r]^{(a, \theta)} & \mathcal{I}_{\mathcal{X}/\mathcal{Y}} }$$
and let $\alpha : (a, \theta) \circ j \to e \circ x$ be any $2$-morphism
witnessing the $2$-commutativity of the diagram (we use $\alpha$ instead
of the letter $\gamma$ used in Definition \ref{definition-fill-in-diagram}).
Note that $f \circ \theta = \text{id}$; we will use this below.
Observe that $e \circ x = (x, \text{id}_x)$ and
$(a, \theta) \circ j = (a \circ j, \theta \star \text{id}_j)$.
Thus we see that $\alpha$ is a $2$-arrow $\alpha : a \circ j \to x$
compatible with $\theta \star \text{id}_j$ and $\text{id}_x$.
Set $y = f \circ x$ and $\beta = \text{id}_{f \circ a}$.
Reading the arguments given in the proof of
Lemma \ref{lemma-cat-dotted-arrows}
backwards, we see that $\theta$ is an automorphism of the
dotted arrow $(a, \alpha, \beta)$ with
$$\gamma : y \circ j \to f \circ x \quad\text{equal to}\quad \text{id}_f \star \alpha : f \circ a \circ j \to f \circ x$$
On the other hand, $\text{id}_a$ is an automorphism too, hence
we conclude $\theta = \text{id}_a$ from the assumption on $f$.
Then we can take as dotted arrow for the displayed diagram above
the morphism $a : \Spec(A) \to \mathcal{X}$ with $2$-morphisms
$(a, \text{id}_a) \circ j \to (x, \text{id}_x)$ given by $\alpha$
and $(a, \theta) \to e \circ a$ given by $\text{id}_a$.

\medskip\noindent
By Lemma \ref{lemma-base-change-existence} any base change of $e$
satisfies the existence part of the valuative criterion.
Since $e$ is representable by algebraic spaces, it suffices to
show that $e$ is universally closed after a base change
by a morphism $I \to \mathcal{I}_{\mathcal{X}/\mathcal{Y}}$
which is surjective and smooth and with $I$ an algebraic space
(see Properties of Stacks, Section
\ref{stacks-properties-section-properties-morphisms}).
This base change $e' : X' \to I'$ is a quasi-compact
morphism of algebraic spaces which
satisfies the existence part of the valuative criterion
and hence is universally closed by
Morphisms of Spaces, Lemma
\ref{spaces-morphisms-lemma-quasi-compact-existence-universally-closed}.
\end{proof}

Comment #2100 by Matthew Emerton on June 22, 2016 a 7:44 pm UTC

Does always a setoid'' meanfor any choice of $\gamma$''? If so, it would be good to make this explicit, since the fact that we have to allow this for every choice of $\gamma$ rather than just our favourite choice seems to be a slightly subtle point.

Comment #2144 by Johan (site) on July 22, 2016 a 2:54 pm UTC

OK, yes you are right. Fixed here.

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